# Continuity And Differentiability

Continuity and differentiability can be easily explained. Consider the function

$f(x)=\begin{cases} x+3 & ,\text{ if } x \leq 0 \\ x & ,\text{ if } x > 0 \end{cases}$

For any point on the Real number line, this function is defined.

It can be seen that the value of the function x = 0 changes suddenly. Following the concepts of limits, we can say that

Right-hand limit $\neq$ Left-hand limit.

It implies that this function is not continuous at x=0.

In layman terms, we can say that we cannot draw the graph of this function until we lift the pen. In simple words, we can say that a function is continuous around a point if we are able to graph it without lifting the pen.

Mathematically, A function is said to be continuous at a point x=a , in its domain if

$\lim \limits_{x \to a}f(x)$ Exists, and

$\lim \limits_{x \to a}f(x)$=f(a)

It implies that if the left hand limit (L.H.L), right hand limit (R.H.L) and the value of the function at x=a exists and these parameters are equal to each other, then the function f is said to be continuous at x=a.

If the function is undefined or does not exist, then we say that the function is discontinuous.

Continuity in open interval (a, b)

f(x) will be continuous in the open interval (a,b) if at any point in the given interval the function is continuous.

Continuity in closed interval [a, b]

A function f(x) is said to be continuous in the closed interval [a,b] if it satisfies the following three conditions.

1) f(x) is be continuous in the open interval (a,b)

2) f(x) is continuous at the point a from right i.e. $\lim\limits_{x \to a}f(x)$=f(a)

3) f(x) is continuous at the point b from left i.e. $\lim\limits_{x \to b}f(x)$=f(b)

 Lets Work Out- Example: Check whether the function $\frac{4x^{2}-1}{2x-1}$ is continuous or not? Solution: At $x=\frac{1}{2}$, the value of denominator is 0. So the function is discontinuous at $x=\frac{1}{2}$.

Differentiability:

f(x) is said to be differentiable at the point x=a if the derivative $f'(a)$ exists at every point in its domain. It is given by

 $f'(a)=\lim\limits_{h \to a}\frac{f(a+h)-f(a)}{h}$

For a function to be differentiable at any point x=a in its domain, it must be continuous at that particular point but vice-versa is not always true.

The derivatives of the basic trigonometric functions are;

1. $\frac{\mathrm{d} }{\mathrm{d} x}(\sin x)= \cos x$
2. $\frac{\mathrm{d} }{\mathrm{d} x}(\cos x)= -\sin x$
3. $\frac{\mathrm{d} }{\mathrm{d} x}(\tan x)= \sec^{2} x$
4. $\frac{\mathrm{d} }{\mathrm{d} x}(\cot x)= – cosec^{2} x$
5. $\frac{\mathrm{d} }{\mathrm{d} x}(\sec x)= \sec x \tan x$
6. $\frac{\mathrm{d} }{\mathrm{d} x}(cosec x)= -cosec x \cot x$
 Lets Work Out- Example: Consider the function $f(x)=(2x-3)^{\frac{1}{5}}$.Discuss its continuity and differentiability at $x= \frac{3}{2}$. Solution: For checking the continuity, we need to check the left hand and right-hand limits and the value of the function at a point x=a. L.H.L. = $\lim\limits_{x \to a^{-}}f(x)= \lim_{x \to \frac{3}{2}}(2x-3)^{\frac{1}{5}}$ $= \left ( 2 \times \frac{3}{2} -3^{\frac{1}{5}} \right )$ = 0 R.H.L. = $\lim\limits_{x \to a^{+}}f(x)= \lim_{x \to \frac{3}{2}}(2x-3)^{\frac{1}{5}}$ $= \left ( 2 \times \frac{3}{2} -3^{\frac{1}{5}} \right )$ = 0 L.H.L = R.H.L = f(a) = 0. Thus the function is continuous at about the point $x= \frac{3}{2}$. Now to check differentiability at the given point, we know $f'(a)=\lim\limits_{h \to a}\frac{f(a+h)-f(a)}{h}$ = $\lim\limits_{h \to 0}\frac{f(\frac{3}{2}+h)-f(\frac{3}{2})}{h}$ = $\lim\limits_{h \to 0}\frac{\left ( 2(\frac{3}{2}+h)-3 \right )^{\frac{1}{5}}- \left ( 2(\frac{3}{2}-3) \right )^{\frac{1}{5}}}{h}$ = $\lim\limits_{h \to 0}\frac{\left ( 3 + h -3 \right )^{\frac{1}{5}}- \left ( 3 -3 \right )^{\frac{1}{5}}}{h}$ = $\lim\limits_{h \to 0}\frac{\left ( h \right )^{\frac{1}{5}}- 0}{h}$ = $\lim\limits_{h \to 0}\frac{ 1}{h^\frac{4}{5}} = \infty$ Thus f is not differentiable at $x= \frac{3}{2}$.. We see that even though the function is continuous but it is not differentiable.